19x2+10x-18=0

Solution for 19x2+10x-18=0 equation:

19x^2+10x-18=0

a = 19; b = 10; c = -18;
Δ = b2-4ac
Δ = 102-4·19·(-18)
Δ = 1468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1468}=\sqrt{4*367}=\sqrt{4}*\sqrt{367}=2\sqrt{367}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{367}}{2*19}=\frac{-10-2\sqrt{367}}{38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{367}}{2*19}=\frac{-10+2\sqrt{367}}{38}
$